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From: Leroy Quet on 24 Jul 2010 15:06
Here is a game for any plural number of players. Start with an n-by-n
grid drawn on paper, where n is larger if there are more players.
Each player has a colored pencil of unique color.

In the first part of the game, players take turns drawing straight
line-segments -- one line-segment each turn. A player can draw a line-
segment from either the edge of the grid at a vertex, or from the end
of another line-segment. (Multiple line-segments can join at one
point.) The line-segments are drawn to any vertex of the grid (either
empty or already occupied by a line segment), such that no line-
segment is drawn through another segment or along another segment or
through a vertex occupied by line-segments. (Although, as I said, a
line-segment may end at a vertex already occupied by another segment.)

After the grid is subdivided into n*(number of players) sections, the
second part of the game begins.
Players take turns filling in sections of the grid with their colored
pencils. After each player fills in n sections, the score is
determined.

Players add up the total area of all the sections in each player's
color, with the area of a grid-square being 1. (This may be tricky
because some sections will most probably have non-integer areas.)

Let a player's total area be m; then the winner is the player where
number of divisors of floor(m) is the SMALLEST.


Thanks,
Leroy Quet
From: Richard Heathfield on 24 Jul 2010 20:17
Leroy Quet wrote:
> Here is a game for any plural number of players. Start with an n-by-n
> grid drawn on paper, where n is larger if there are more players.
> Each player has a colored pencil of unique color.

That puts an upper limit on the number of players, since the human eye
can only distinguish approximately 10,000,000 colours.

<snip>

> Players add up the total area of all the sections in each player's
> color, with the area of a grid-square being 1. (This may be tricky
> because some sections will most probably have non-integer areas.)

Not so tricky. Count the interior vertices (bounded by, but not
coincident with, the borders of a given colour area). Add half the
number of vertices that /do/ coincide with the border, including any
that are collinear. Subtract 1. The result will be either integral or
integral + a half. If there are any "holes", add the number of holes to
the sum. Your final result is the area of the coloured section. (Pick's
Theorem.)

<snip>

--
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
"Usenet is a strange place" - dmr 29 July 1999
Sig line vacant - apply within
From: Leroy Quet on 25 Jul 2010 07:32


Richard Heathfield wrote:
> Leroy Quet wrote:
> > Here is a game for any plural number of players. Start with an n-by-n
> > grid drawn on paper, where n is larger if there are more players.
> > Each player has a colored pencil of unique color.
>
> That puts an upper limit on the number of players, since the human eye
> can only distinguish approximately 10,000,000 colours.
>

In the case there are more than 10 million players, players should
mark each section with a symbol or with their name. Since many players
will have the same names, in this case, follow each name with a
number: Like "James Albert Eugene 8165237", say.
Still, the grid will have to be less than the size of the universe,
and so this limits the number of players, since, one, a grid too small
can't be subdivided into enough sections to let every player claim at
least one section, and two, there are only so many potential players
in the universe anyway.



> <snip>
>
> > Players add up the total area of all the sections in each player's
> > color, with the area of a grid-square being 1. (This may be tricky
> > because some sections will most probably have non-integer areas.)
>
> Not so tricky. Count the interior vertices (bounded by, but not
> coincident with, the borders of a given colour area). Add half the
> number of vertices that /do/ coincide with the border, including any
> that are collinear. Subtract 1. The result will be either integral or
> integral + a half. If there are any "holes", add the number of holes to
> the sum. Your final result is the area of the coloured section. (Pick's
> Theorem.)


Ah, yes, Pick's theorem. I forgot about that. Not so tricky, then.

Thanks,
Leroy Quet

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